I'm working with finding roots of polynomial equations with degrees of 3 or higher. I have the equation
r(x)=x^4-6x^3+12x^2=6x-13
I used a graphing calculator to find the real roots of 1,-1
Then I did synthetic using -1, and I ended up with the equation
x^3-6x^2+2x-13
How do I get this to a quadratic so that I can find the imaginary roots? A friend suggested grouping, but I'm not real good at grouping, and I couldn't figure out how to factor it or group it. Any help would be appreciated.
Thanks,
Josh
To find the imaginary roots of the polynomial equation x^3 - 6x^2 + 2x - 13, you can try grouping the terms. When grouping, you break up the polynomial into two groups and factor out the common terms.
Let's rearrange the equation:
x^3 - 6x^2 + 2x - 13 = 0
Now, try grouping the terms:
(x^3 - 6x^2) + (2x - 13) = 0
Now, look for common factors in each group:
x^2(x - 6) + (2x - 13) = 0
Since there aren't any common factors between the two groups, let's try factoring by grouping in a different way.
(x^2 + 2x) + (-6x - 13) = 0
In this case, you can factor out common terms:
x(x + 2) - 13(6x + 13) = 0
Now, we have grouped the terms properly. To get this into quadratic form, we need to factor out the common terms.
x(x + 2) - 13(6x + 13) = 0
Applying the distributive property:
x^2 + 2x - 78x - 169 = 0
Combining like terms:
x^2 - 76x - 169 = 0
Now, you have transformed the original polynomial equation into quadratic form: x^2 - 76x - 169 = 0. From here, you can use various methods such as factoring, completing the square, or the quadratic formula to find the roots of this quadratic equation, which will give you the imaginary roots of the original polynomial equation.